Answer
$f(x)$ is not differentiable on the interval [-1,1] since it is not differentiable at x = 0 due to cusp behavior.
Work Step by Step
For Rolle's Theorem to work, $f(x)$ must be differentiable over the closed interval. This is not true for $f(x) = \sqrt{(2-x^{2/3})^3}$ since it exhibits cusp-like behavior at x = 0. Because $f(x)$ is not differentiable at all points between the interval, Rolle's Theorem fails. The graph below depicts $f(x)$ and shows how $f(x)$ is not differentiable at x = 0.
